Percent yield measures how efficient a reaction and its work-up were. The theoretical yield is what stoichiometry promises if every molecule of the limiting reagent turns into product and nothing is lost; the actual yield is what you really weigh at the end. Their ratio, multiplied by 100, is the percent yield.
In teaching labs, yields of 70–90% are typical for a careful single step, while multi-step research syntheses often accept much less. Low yields point to incomplete reaction, side reactions, unfavourable equilibria or losses during filtration, extraction and recrystallisation. A yield over 100% is a warning sign rather than a triumph: the product is almost always still wet with solvent or water, or contaminated with by-products or unreacted starting material.
Both yields must be in the same units. Enter two masses, or two amounts in moles; if your theoretical yield is in moles but you weighed your product, choose the mixed mode and give the molar mass of the product so the calculator can convert. You also get the amount of product lost and a rating band.
How to use the Percent Yield Calculator
- Enter the yields as
- Actual yield (mass)
- Theoretical yield (mass)
- Actual yield (moles)
- Theoretical yield (moles)
- Molar mass of the product
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
The calculator divides actual by theoretical yield (after converting the theoretical moles to grams with the molar mass in the mixed mode), multiplies by 100, and subtracts to find the amount lost.
Worked example
A student makes aspirin and the stoichiometry predicts 2.61 g. After recrystallising and drying, 2.15 g is weighed. Percent yield = 2.15 ÷ 2.61 × 100 = 82.4%; 0.46 g (17.6%) was lost – a good result.
If the theoretical yield had been worked out as 0.01448 mol, the mixed mode first converts it: 0.01448 × 180.16 = 2.609 g, giving 2.15 ÷ 2.609 = 82.4% as well.
Where yield is lost
- Incomplete reaction or an equilibrium that does not lie fully on the product side.
- Side reactions that turn reactants into by-products.
- Product left in the mother liquor after recrystallisation or on glassware and filter paper.
- Decomposition during heating or drying.
- Weighing errors and impure or wet starting materials.
Inputs explained
| Field | What to enter |
|---|---|
| Enter the yields as | Choose one: Masses, Amounts in moles, Actual mass and theoretical moles (with molar mass) |
| Actual yield (mass) | Units: g, mg, µg, kg, oz, lb |
| Theoretical yield (mass) | Units: g, mg, µg, kg, oz, lb |
| Actual yield (moles) | Units: mol, mmol, µmol, nmol, kmol |
| Theoretical yield (moles) | Units: mol, mmol, µmol, nmol, kmol |
| Molar mass of the product | Units: g/mol, kg/mol, Da (u), kDa |
What each result means
- Percent yield = ([Fraction])×100
- Theoretical yield as mass = [Theoretical yield (moles)]×[Molar mass of the product]
- Product lost (theoretical − actual) = [Theoretical yield (moles)] – [Actual yield (moles)]
- Percent lost = (1 – [Fraction])×100
Example values
How the main result changes when Actual yield changes and everything else stays at the example values above:
| Actual yield (mass) (g) | Percent yield (%) |
|---|---|
| 1.08 | 41.19 |
| 1.61 | 61.78 |
| 2.15 | 82.38 |
| 2.69 | 102.97 |
| 3.22 | 123.56 |
| 4.3 | 164.75 |
Formula
Percent yield = actual yield ÷ theoretical yield × 100% · mixed units: % = m_actual ÷ (n_theoretical × M) × 100%
Worked example
Inputs — Enter the yields as: Masses; Actual yield (mass): 2.15 g; Theoretical yield (mass): 2.61 g.
Results — Percent yield = 82.38 %; Product lost (theoretical − actual) = 0.46 g; Percent lost = 17.62 %; Assessment = Good (70–90%).
Key terms
- Percent yield
- Actual yield as a percentage of the theoretical yield.
- Theoretical yield
- The maximum product predicted from the limiting reagent.
- Actual yield
- The amount of product actually isolated.
- Limiting reagent
- The reactant that runs out first and caps the amount of product.
Frequently asked questions
How do you calculate percent yield?
Divide the actual yield by the theoretical yield and multiply by 100. For 5.58 g obtained out of a possible 6.54 g, the percent yield is 85.3%.
What is a good percent yield?
In a teaching lab 70% or more is usually considered good and 90% or more excellent. In multi-step research synthesis, 50% for a difficult step can be acceptable.
Why is my percent yield over 100%?
The product still contains solvent or water, or impurities such as unreacted starting material. Dry to constant mass and check purity (for example by melting point).
Do both yields need the same units?
Yes – both in grams or both in moles. If the theoretical yield is in moles and the actual in grams, multiply the moles by the molar mass first (the mixed mode does this).
What is the difference between percent yield and atom economy?
Percent yield measures how much of the possible product you collected; atom economy measures what fraction of the reactant atoms end up in the desired product even at 100% yield.
How do I find the actual yield from the percent yield?
Rearrange: actual = theoretical × percent ÷ 100. 70% of 15 g is 10.5 g.